What is the limit of #ln(2x)-ln(1+x)# as x goes to infinity?

2 Answers
Gió
Oct 6, 2015

I found #ln(2)#

Explanation:

Let us use a rule of the logs:
#lim_(x->oo)[ln(2x)-ln(1+x)]=#
#lim_(x->oo)[ln((2x)/(1+x))]=# collect #x#:
#lim_(x->oo)[ln((2cancel(x))/(cancel(x)(1/x+1)))]=#
as #x->oo# then #1/x->0#
#=ln(2)#

Jim H
Oct 6, 2015

Use the properties of #ln# to rewrite as a single #ln# then use continuity of #ln#.

Explanation:

Rewrite as #lnu#, tghen use the following.

Because #ln# is continuous on #(0,oo)#, we have

if #lim_(xrarroo)u = L# for some number #L#, then

#lim_(xrarroo) lnu = ln(lim_(xrarroo)u)#.

If you just want the answer it is #ln2#.

Related questions
See all questions in Limits at Infinity and Horizontal Asymptotes
Impact of this question
2847 views around the world
You can reuse this answer
Creative Commons License